Q4001 Thermodynamics of Materials
Midterm Exam. Part I
Antonio Osamu Katagiri Tanaka - A01212611@itesm.mx
from IPython.display import display, Image
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Q4001 Thermodynamics of Materials
Midterm Exam. Part II
Antonio Osamu Katagiri Tanaka - A01212611@itesm.mx
"I state that the work I am submitting for this exam is solely my own and that I have not shared it. I have not used information coming from any uncited source in my answers."
display(Image(filename='./img/part2_directions.jpg'));
Q 6
The 1st law gives, $$ dU = T dS - P dV $$ $$ dS = \frac{dU + P dV}{T} $$ $$ dS = \frac{dU}{T} + \frac{P dV}{T} $$ $$ dS = \left[\frac{1}{T} \left(\frac{\partial U}{\partial V}\right)_T + \frac{P}{T}\right] dV + \frac{1}{T} \left(\frac{\partial U}{\partial T}\right)_V dT $$
$$ \left(\frac{\partial U}{\partial V}\right)_T = \frac{B T^n}{V} \text{ and } \left(\frac{\partial U}{\partial T}\right)_V = f'(T) + B n T^{n - 1} ln\left(\frac{V}{V_0}\right) $$
So, $$ dS = \left[\frac{1}{T} \frac{B T^n}{V} + \frac{\frac{a T^3}{V}}{T}\right] dV + \frac{1}{T} \left[f'(T) + B n T^{n - 1} ln\left(\frac{V}{V_0}\right)\right] dT $$ $$ dS = \frac{B T^{n-1} + a T^2}{V} dV + \left[\frac{f'(T)}{T} + n B T^{n-2} ln\left(\frac{V}{V_0}\right)\right] dT $$ $$ \frac{\partial}{\partial T} \left(\frac{B T^{n-1} + a T^2}{V}\right) = \frac{\partial}{\partial V} \left(\frac{f'(T)}{T} + n B T^{n-2} ln\left(\frac{V}{V_0}\right)\right) $$ $$ \frac{2 a T + B (n - 1) T^{n-2}}{V} = \frac{B n T^{n-2}}{V} $$ $$ 2 a T = B b T^{n - 2} - B (n-1) T^{n-2} $$ $$ 2 a T = B T^{n-2} $$
[1] Lim, Y.-K. (2019). Problems and Solutions on Thermodynamics and Statistical Mechanics. Journal of Chemical Information and Modeling, 53(9), 1689–1699. https://doi.org/10.1017/CBO9781107415324.004
Q 7
@ constant volume $$ dU = T dS - P dV $$ $$ \left(\frac{\partial S}{\partial U}\right)_V = \frac{1}{T} $$
From: $$ S = \frac{1}{2} \left[\sigma + 5 R lnU + 2 R lnV\right] $$ $$ \left(\frac{\partial S}{\partial U}\right)_V = \frac{5 R}{2 U} $$ $$ U = \frac{5}{2} R T $$
From the $c_v$ definition,
From the $c_p$ definition,
From Problem 9, $$ c_p - c_v = \left(\frac{\partial V}{dT}\right)_P \left[ \left(\frac{\partial U}{\partial V}\right)_T + P \right] $$
with: $$ \alpha = \frac{1}{v} \left(\frac{\partial V}{\partial T}\right)_P $$ $$ P = -\left(\frac{\partial A}{\partial V}\right)_T $$
$$ c_p - c_v = \alpha V \left[ \left(\frac{\partial U}{\partial V}\right)_T - \left(\frac{\partial A}{\partial V}\right)_T \right] $$$$ A = U - TS $$$$ \left(\frac{\partial A}{\partial V}\right)_T = \left(\frac{\partial U}{\partial V}\right)_T - T \left(\frac{\partial S}{\partial V}\right)_T $$$$ - \left(\frac{\partial A}{\partial V}\right)_T = - \left(\frac{\partial U}{\partial V}\right)_T + T \left(\frac{\partial S}{\partial V}\right)_T $$
$$ c_p - c_v = \alpha V \left[ T \left(\frac{\partial S}{\partial V}\right)_T \right] = T \alpha V \left(\frac{\partial P}{\partial T}\right)_V $$
$$ \left(\frac{\partial P}{\partial T}\right)_V = -\left(\frac{\partial P}{\partial V}\right)_T \left(\frac{\partial V}{\partial T}\right)_P $$
$$ c_p - c_v = T \alpha V \left[-\left(\frac{\partial P}{\partial V}\right)_T \left(\frac{\partial V}{\partial T}\right)_P\right] $$$$ c_p - c_v = (T \alpha V) \left(\frac{1}{V k_T}\right) (V \alpha) $$$$ c_p - c_v = \frac{V T \alpha^2}{k_T} $$$$ \frac{\alpha}{k_T} = \frac{R}{T V \alpha} $$
Since $$ \frac{\alpha}{k_T} = \frac{R}{V} = \frac{R}{T V \alpha} $$
$$ k_T = \frac{V T \alpha^2}{c_p - c_v} $$$$ k_T = \frac{V T \alpha^2}{R} $$
[1] Lim, Y.-K. (2019). Problems and Solutions on Thermodynamics and Statistical Mechanics. Journal of Chemical Information and Modeling, 53(9), 1689–1699. https://doi.org/10.1017/CBO9781107415324.004
[2] R. Gaskell, D., & E. Laughlin, D. (2018). Introcuction to the Thermodynamics of Materials. (C. Press, Ed.). Taylor & Francis Group.
Q 8
$$ dU = T dS + H dM $$gives: $$ \left(\frac{\partial T}{\partial M}\right)_S = \left(\frac{\partial H}{\partial S}\right)_M $$
From a Maxwell relation: $$ \left(\frac{\partial T}{\partial S}\right)_H \left(\frac{\partial M}{\partial H}\right)_H = -1 $$
Also: $$ c_H = T \left(\frac{\partial S}{\partial T}\right)_H $$ $$ \left(\frac{\partial T}{\partial S}\right)_H = \frac{T}{c_H} $$
Recall: $$ M = \frac{a H}{T} $$ $$ \left(\frac{\partial M}{\partial T}\right)_H = -\frac{a H}{T^2} $$
$$ \left(\frac{\partial T}{\partial H}\right)_S = \left(\frac{\partial T}{\partial S}\right) \left(\frac{\partial S}{\partial H}\right) = \left(\frac{\partial H}{\partial S}\right) \left(\frac{\partial T}{\partial H}\right) \cdot \left(\frac{\partial H}{\partial T}\right) \left(\frac{\partial M}{\partial H}\right) \cdot \left(\frac{\partial T}{\partial S}\right) \left(\frac{\partial M}{\partial H}\right) $$$$ \left(\frac{\partial T}{\partial H}\right)_S = \left(\frac{\partial T}{\partial S}\right)_H \left(\frac{\partial M}{\partial T}\right)_H (-1) $$$$ \left(\frac{\partial T}{\partial H}\right)_S = -\frac{T}{c_H} \frac{a H}{T^2} $$$$ \left(\frac{\partial T}{\partial H}\right)_S = \frac{a T H}{b} $$
$$ \int \frac{a H T}{b} \, dH = \frac{a H^2 T}{2 b} = ln\left(\frac{T}{T_f}\right)$$
[1] Lim, Y.-K. (2019). Problems and Solutions on Thermodynamics and Statistical Mechanics. Journal of Chemical Information and Modeling, 53(9), 1689–1699. https://doi.org/10.1017/CBO9781107415324.004
[3] R.L., C. (1986). Thermodynamics, Magnetoche(Springer), 3–4. Retrieved from https://link.springer.com/chapter/10.1007%2F978-3-642-70733-9_3
Q 9
Let's demonstrate: $\bar{C_P} - \bar{C_V} = \left[ P + \left(\frac{\partial U}{\partial V} \right)_T \right] \left(\frac{\partial V}{\partial T}\right)_P$
For a constant volume process ($w = 0$), the 1st Law gives: $$ dU = \delta q_v $$
From the definition of enthalpy: $$ dH = dU + P dV + V dP $$ @ constant pressure: $$ dH = dU + P dV $$ $$ dH = \delta q_p $$
From the definition of $c_v$ and $c_p$, $$ c_v = \left(\frac{\delta q}{dT}\right)_v = \left(\frac{\partial U}{dT}\right)_v \text{ , as } dU = c_v dT$$ $$ c_p = \left(\frac{\delta q}{dT}\right)_p = \left(\frac{\partial H}{dT}\right)_p \text{ , as } dH = c_p dT$$
If the process is carried out at constant volume, all the thermal energy is used to raise the temperature of the system.
If the process is carried out at constant pressure, some thermal energy is used to raise the temperature of the system and some is used to provide the work required to expand the system. $$ \frac{P dV}{dT} = P \left(\frac{\partial V}{\partial T}\right)_p \text{ is the work of expansion} $$
So, it's expected $c_p$ to be greater than $c_v$. Therefore:
$$ c_p - c_v = P \left(\frac{\partial V}{\partial T}\right)_p $$From the definition of enthalpy: $$ H = U + PV $$ $$ dH = dU + P dV + V dP $$ $$ dH = dU + P dV \text{ @ constant pressure} $$ $$ \left(\frac{\partial H}{dT}\right)_p = \left(\frac{\partial U}{dT}\right)_p + P \left(\frac{\partial V}{dT}\right)_p $$
So: $$ c_p = \left(\frac{\partial H}{dT}\right)_p = \left(\frac{\partial U}{dT}\right)_p + P \left(\frac{\partial V}{dT}\right)_p $$ And: $$ c_v = \left(\frac{\partial U}{dT}\right)_v $$ For any gas
$$ c_p - c_v = \left(\frac{\partial U}{dT}\right)_p + P \left(\frac{\partial V}{dT}\right)_p - \left(\frac{\partial U}{dT}\right)_v $$
$$ U = U(V, T) $$$$ dU = \left(\frac{\partial U}{\partial V}\right)_T dV + \left(\frac{\partial U}{\partial T}\right)_V dT $$$$ \left(\frac{\partial U}{\partial T}\right)_P = \left(\frac{\partial U}{\partial V}\right)_T \left(\frac{\partial V}{\partial T}\right)_P + \left(\frac{\partial U}{\partial T}\right)_V (1) $$
So, $$ c_p - c_v = \left(\frac{\partial U}{\partial V}\right)_T \left(\frac{\partial V}{\partial T}\right)_P + \left(\frac{\partial U}{\partial T}\right)_V + P \left(\frac{\partial V}{dT}\right)_P - \left(\frac{\partial U}{dT}\right)_V $$ $$ c_p - c_v = \left(\frac{\partial U}{\partial V}\right)_T \left(\frac{\partial V}{\partial T}\right)_P + P \left(\frac{\partial V}{dT}\right)_P $$
Let's demonstrate: $P + \left(\frac{\partial U}{\partial V}\right)_T = T \left(\frac{\partial P}{\partial T}\right)_V$
For one mole in a closed system: $$ dU = T dS - P dV $$ $$ \left(\frac{\partial U}{\partial V}\right)_T = T \left(\frac{\partial S}{\partial V}\right)_T - P (1) $$ $$ \text{using the Maxwell relation: } \left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V $$ $$ \left(\frac{\partial U}{\partial V}\right)_T = T \left(\frac{\partial P}{\partial T}\right)_V - P $$
Let's calculate $\bar{C_P} - \bar{C_V}$ for a van der Waals gas.
$$ dS = \left(\frac{\partial S}{\partial T}\right)_V dT + \left(\frac{\partial S}{\partial V}\right)_T dV $$Differentiate with respect to T (P constant) $$ \left(\frac{\partial S}{\partial T}\right)_P = \left(\frac{\partial S}{\partial T}\right)_V + \left(\frac{\partial S}{\partial V}\right)_T \left(\frac{\partial V}{\partial T}\right)_P $$
From the definitios of the heat capacities: $$ c_v = T \left(\frac{\partial S}{\partial T}\right)_V \text{ and } c_p = T \left(\frac{\partial S}{\partial T}\right)_P $$ $$ \frac{c_p - c_v}{T} = \left(\frac{\partial S}{\partial T}\right)_P - \left(\frac{\partial S}{\partial T}\right)_V $$ $$ \frac{c_p - c_v}{T} = \left(\frac{\partial S}{\partial T}\right)_V + \left(\frac{\partial S}{\partial V}\right)_T \left(\frac{\partial V}{\partial T}\right)_P - \left(\frac{\partial S}{\partial T}\right)_V $$ $$ \frac{c_p - c_v}{T} = \left(\frac{\partial S}{\partial V}\right)_T \left(\frac{\partial V}{\partial T}\right)_P $$
Maxwell relationship: $$ \left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V $$
The Cycle Rule: $$ \left(\frac{\partial P}{\partial T}\right)_V \left(\frac{\partial T}{\partial V}\right)_P \left(\frac{\partial V}{\partial P}\right)_T = -1 $$ $$ \left(\frac{\partial V}{\partial T}\right)_P = -\left(\frac{\partial P}{\partial T}\right)_V \left(\frac{\partial V}{\partial P}\right)_T $$
So: $$ \frac{c_p - c_v}{T} = \left(\frac{\partial P}{\partial T}\right)_V \left[-\left(\frac{\partial P}{\partial T}\right)_V \left(\frac{\partial V}{\partial P}\right)_T\right] $$ $$ c_p - c_v = -T \frac{{\left(\frac{\partial P}{\partial T}\right)_V}^{2}}{\left(\frac{\partial P}{\partial V}\right)_T} $$
For a van der Waals gas, $$ P = \frac{R T}{V - b} - \frac{a^2}{V^2} $$ $$ \left(\frac{\partial P}{\partial T}\right)_V = \frac{R}{V - b} \text{ and } \left(\frac{\partial P}{\partial V}\right)_T = \frac{2 a^2}{V^3} - \frac{R T}{{(V - b)}^2} $$
Thus, $$ c_p - c_v = -T \frac{{\left(\frac{R}{V - b}\right)}^{2}}{\frac{2 a^2}{V^3} - \frac{R T}{{(V - b)}^2}} $$
Let's compare $c_p - c_v$ for $He$ and $CO_2$
def Cp_min_Cv(T, V, R, a, b):
numer = (R**2)*T*(V**3);
denom = -2*(a**2)*(V - b)**2 + R*T*(V**3);
return numer / denom;
# constants
T = 200; #K
V = 0.2; #L/mol
R = 0.082057; #L atm/K mol
# for He
a = 0.0346; #atm L^2/mol^2
b = 0.0238; #L/mol
He_Cp_min_Cv = Cp_min_Cv(T, V, R, a, b)
# for CO2
a = 3.64; #atm L^2/mol^2
b = 0.0427; #L/mol
CO2_Cp_min_Cv = Cp_min_Cv(T, V, R, a, b)
print("gas Cp_min_Cv")
print("---------------")
print("{} {:>5.3f} {}".format("He ", He_Cp_min_Cv, "L atm/K mol"))
print("{} {:>5.3f} {}".format("CO2", CO2_Cp_min_Cv, "L atm/K mol"))
[1] R. Gaskell, D., & E. Laughlin, D. (2018). Introcuction to the Thermodynamics of Materials. (C. Press, Ed.). Taylor & Francis Group.
[2] Berberan-santos, M. N. (2008). The van der Waals equation : analytical and approximate solutions, 43(4), 1437–1457. https://doi.org/10.1007/s10910-007-9272-4
Q 10
$$ \frac{T_H - T_C}{L} \text{ is the bar temperature gradient} $$$$ T_x = T_C + \frac{(T_H - T_C)x}{L} \text{ is the temperature at the cross section at distance x} $$Since the process is isobaric ans adiabatic, $$ \int_0^L \rho c_p \left[T_C + \frac{(T_H - T_C)x}{L} - T_f\right] \, dx = 0 $$ $$ \frac{1}{2} L \rho c_p T_C - L \rho c_p T_f + \frac{1}{2} L \rho c_p T_H = 0 $$ $$ \frac{1}{2} T_C - T_f + \frac{1}{2} T_H = 0 $$
Let's calculate the change in entropy $$ \Delta S = c_p \rho A \int_0^L \, dx \int_{T_x}^{T_f} \, \frac{dT}{T} $$ $$ \Delta S = c_p \rho A L \left[ln\left(\frac{T_H + T_C}{2}\right) - ln\left(T_C + \frac{(T_H - T_C)x}{L}\right)\right] $$
[1] Lim, Y.-K. (2019). Problems and Solutions on Thermodynamics and Statistical Mechanics. Journal of Chemical Information and Modeling, 53(9), 1689–1699. https://doi.org/10.1017/CBO9781107415324.004